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Let $\alpha = (1 6 2) (4 5)$ and $\beta = (1 2 3 6)$ be permutations in $S_ {6}$. Calculate the product $\beta \alpha$. Answer with a product of disjunked cycles.

I don't know how to do it, but maybe something like this;

Step 1: I think that if we look at $\alpha(1)$, we see that in $\beta$ that that answers to 2. And 2 in $\alpha$ is equal to 1, so that is our first cycle, (1 2).

Step two: And then we look at $\alpha(2)$ and that is in $\beta$=1 and that is in $\alpha$ = 6.. So that's another cycle?

Step three: We're no looking at $\alpha(3)$ with is in $\beta$=6. So there's another cycle (3 6) and we don't have a 3 in $\alpha$ so this is another cycle..

Step four: $\alpha(4), \alpha(5)$ dosen't have any 'friends' in $\beta$ so this is the other cycle (4)(5).

Therefore, the answer should be (1 2)(2 6)(3)(4)(5)

hhmmm.. But it's wrong. Someone help me out? :)

soetirl13
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1 Answers1

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So we have: $α=(162)(45)$

$β=(1236)$

$\alpha \beta =(1)(23)(45)(6)=(23)(45)$

You do this: you look at where 1 get map to by $\beta$, it gets map to 2, then you look at where 2 gets map to by $\alpha$, it gets map to 1. So 1 goes to 1 by $\alpha\beta$. Then you look at where 2 gets map to by $\beta$, it gets map to 3 and then you look at where 3 gets map to by $\alpha$, it is being fixed. so 2 gets map to 3 by $\alpha\beta$. Then you look at where 3 gets map to by $\beta$, it gets map to 6, and you look at where 6 gets map to by $\alpha$, it gets map to 2. So 3 gets map to 2 by $\alpha\beta$ and you close the cycle. You keep doing this until you know where every numbers gets map to to get a disjoint union of cycles. And then you can remove cycles of length 1 because they are redundant as you can just say numbers not present in your cycle decomposition gets map to themselves.

user614287
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  • Ok, thanks:) But the question was $\beta \alpha$ .., – soetirl13 Nov 25 '18 at 16:19
  • Ok then the solution is $(1)(2)(36)(45)=(36)(45)$. – user614287 Nov 25 '18 at 16:22
  • how did you do that? I thought of this: 1 goes to $\alpha$ it gets to 6 in $\beta$, then 6 goes to 1. So 1 goes to 6 in \alpha\beta.

    Then we look at where 6 map to by $\beta$ it goes to 1 in $\alpha \beta$ and 1 in $\alpha$ goes to 2, and 2 in $\beta$ goes to 3. We don’t have any 3 in $\alpha$, same with 4 and 5 in alpha, it dosen’t go anywhere in $\beta$

    So answer is: (16)(23) nah?? wrong :/

     – soetirl13 Nov 25 '18 at 16:28
  • 1 goes to 6 in $\alpha$ and 6 goes to 1 in $\beta$ so 1 goes to 1 in $\beta \alpha$ not 6. Think $\alpha and \beta as bijective functions and you are composing them. – user614287 Nov 25 '18 at 16:35
  • but is it so, that for example; $(a b c d)$ , $a->b, b->c, c->d, d->a$ right? – soetirl13 Nov 25 '18 at 16:37
  • yes but you follow the arrow once for each function. not all the way till the end. there is no end. – user614287 Nov 25 '18 at 16:39
  • Okej, think I got it. :) Thanks – soetirl13 Nov 25 '18 at 16:43
  • Np, when I wrote function, I meant permutations. – user614287 Nov 25 '18 at 16:45